James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

394 Chapter 5 Integrals
Intuitively, we therefore expect that
tsx 1 hd 2 tsxd
t9sxd − lim
− f sxd
h l 0 h
The fact that this is true, even when f is not necessarily positive, is the first part of the
Fun damental Theorem of Calculus.
We abbreviate the name of this theorem
as FTC1. In words, it says that the
derivative of a definite integral with
respect to its upper limit is the integrand
evaluated at the upper limit.
The Fundamental Theorem of Calculus, Part 1 If f is continuous on fa, bg, then
the function t defined by
tsxd − y x
f std dt
a
a < x < b
is continuous on fa, bg and differentiable on sa, bd, and t9sxd − f sxd.
Proof If x and x 1 h are in sa, bd, then
tsx 1 hd 2 tsxd − y x1h
f std dt 2 y x
f std dt
a
−Sy x
a
− y x1h
f std dt
x
a
f std dt 1 y x1h
f std dtD 2 y x
f std dt (by Property 5)
x
a
y
and so, for h ± 0,
y=ƒ
2
tsx 1 hd 2 tsxd
h
− 1 y x1h
f std dt
h
x
m
M
For now let’s assume that h . 0. Since f is continuous on fx, x 1 hg, the Extreme
Value Theorem says that there are numbers u and v in fx, x 1 hg such that f sud − m
and f svd − M, where m and M are the absolute minimum and maximum values of f on
fx, x 1 hg. (See Figure 6.)
By Property 8 of integrals, we have
0
FIGURE 6
x u √=x+h
x
mh < y x1h
f std dt < Mh
x
that is,
f sudh < y x1h
f std dt < f svdh
x
Since h . 0, we can divide this inequality by h:
f sud < 1 y x1h
f std dt < f svd
h
x
Now we use Equation 2 to replace the middle part of this inequality:
3
f sud <
tsx 1 hd 2 tsxd
h
< f svd
TEc Module 5.3 provides visual
evidence for FTC1.
Inequality 3 can be proved in a similar manner for the case where h , 0. (See Exercise
77.)
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